Periodic Table Formulae

Elements exist in different forms depending on their group. Noble gases are monatomic, meaning they exist as single, unbonded atoms that do not form ions in standard contexts. In contrast, halogens and a few other non-metals are diatomic, existing as molecules made of two atoms chemically bonded together ($\text{H}_2, \text{N}_2, \text{O}_2, \text{F}_2, \text{Cl}_2, \text{Br}_2, \text{I}_2$).

Transition metals are found in the central block of the Periodic Table. They have high densities, variable oxidation states (can form ions with different charges), and typically form coloured compounds. Common examples include iron ($\text{Fe}$), copper ($\text{Cu}$), and zinc ($\text{Zn}$). Other common elements frequently appearing in OCR papers include silver ($\text{Ag}$) and the Group 2 metal barium ($\text{Ba}$).

Deducing Chemical Formulae

A chemical formula represents a substance using symbols and subscripts to show the exact ratio of atoms. To deduce the formula of an ionic compound, you must know the valency of each element. Valency is the "combining power" of an atom, linking directly to the electrons in its outer shell and its Periodic Table group.

• Group 1: Valency 1
• Group 2: Valency 2
• Group 3: Valency 3
• Group 5: Valency 3
• Group 6: Valency 2
• Group 7: Valency 1

Transition metals often have their valency indicated by a Roman numeral (e.g., Copper(II) means $\text{Cu}^{2+}$). You must also memorise common polyatomic ions: hydroxide ($\text{OH}^-$), nitrate ($\text{NO}_3^-$), sulfate ($\text{SO}_4^{2-}$), carbonate ($\text{CO}_3^{2-}$), and ammonium ($\text{NH}_4^+$).

Worked Example: Deducing Formulae

Calculate the chemical formula for aluminium sulfate using the cross-over method.

Step 1: Write the symbols for the elements or polyatomic ions.

• $\text{Al}$ and $\text{SO}_4$

Step 2: Write the valency (charge magnitude) below each symbol.

• $\text{Al}$ is in Group 3, so its valency is 3.
• Sulfate ($\text{SO}_4^{2-}$) has a valency of 2.

Step 3: Cross over the valency numbers to become the subscripts for the opposite symbol.

• The 2 goes to $\text{Al}$, becoming $\text{Al}_2$.
• The 3 goes to $\text{SO}_4$, becoming $(\text{SO}_4)_3$.

Step 4: Write the final simplified formula. Brackets are essential because there is more than one polyatomic ion.

• $\text{Al}_2(\text{SO}_4)_3$

The Law of Conservation of Mass

The Law of Conservation of Mass states that mass is never lost or gained in a chemical reaction. Atoms are simply rearranged to form new products, so the total mass of the reactants always exactly equals the total mass of the products.

In a closed system, no substances can enter or leave, meaning the measured mass remains visibly constant throughout the reaction. However, in an open system, gases can enter from the atmosphere or escape into the surroundings. This can cause the measured mass on a balance to apparently increase (if a gas reacts with a solid) or decrease (if a product gas escapes).

Writing and Balancing Equations

To obey the conservation of mass, we write a balanced chemical equation. This ensures the number of atoms for each element is identical on both sides of the arrow.

We achieve this by adding stoichiometric coefficients — large numbers placed in front of a formula. You must never change the small subscript numbers inside a chemical formula, as this would alter the molecular structure and completely change the identity of the substance.

Worked Example: Formulae and Balancing

Write a balanced chemical equation for the reaction between lithium and oxygen to form lithium oxide.

Step 1: Deduce the formulae of the reactants and products using valency.

• Lithium is a Group 1 metal: $\text{Li}$
• Oxygen is a diatomic gas: $\text{O}_2$
• Lithium oxide combines $\text{Li}$ (valency 1) and $\text{O}$ (valency 2). Crossing over gives $\text{Li}_2\text{O}$.

Step 2: Write the unbalanced equation.

• $\text{Li} + \text{O}_2 \rightarrow \text{Li}_2\text{O}$

Step 3: Count the atoms on both sides.

• Left: 1 $\text{Li}$, 2 $\text{O}$
• Right: 2 $\text{Li}$, 1 $\text{O}$

Step 4: Balance the oxygen by placing a coefficient of 2 in front of lithium oxide.

• $\text{Li} + \text{O}_2 \rightarrow 2\text{Li}_2\text{O}$
• (The right side now has 4 $\text{Li}$ and 2 $\text{O}$)

Step 5: Balance the lithium by placing a coefficient of 4 in front of lithium.

• $4\text{Li} + \text{O}_2 \rightarrow 2\text{Li}_2\text{O}$

State Symbols

State symbols are small letters in brackets placed after a chemical formula to indicate its physical state at the time of the reaction. The four symbols are $\text{(s)}$ for solid, $\text{(l)}$ for pure liquid, $\text{(g)}$ for gas, and $\text{(aq)}$ for aqueous (dissolved in water).

A common reaction involving aqueous solutions can produce a precipitate, which is an insoluble solid formed when two solutions are mixed.

In this equation, the $\text{(s)}$ identifies silver chloride as the solid precipitate, while the other substances remain fully dissolved in the water.